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Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°
Answer:
S varies partly directly as M and Q.
S=C.
S=KMQ+C.
For the first one...
speed=80,m=220,Q=30.
80=K20×30+C.
80=600K+C......(I).equation one.
For the second one....
speed=60,m=300,Q=40.
60=K300×40+C.
60=12000K+C.....(ii). equation two.
Minus eqtn(I) from eqtn(ii).
80=600K+C.
- 60=12000K+C.
K=0.01754~0.018.
Substitute K=0.018 into eqtn(I).
80=600K+C
80=600×0.018+C.
80=10.8+C.
C=80-10.8=69.2.
The relation is S=0.018MQ+69.2
when speed is 100 and mass is 250 find the volume.
100=0.018×250×Q+69.2.
100=4.5Q+69.2.
4.5Q=100-69.2
4.5Q=30.8.
Q=30.8/4.5.
Q=6.8~7litres.
If 100 miles = 1 in.
650 = 6.5 x 100 and 6.5 x 1 = 6.5
so, if the journey is 650 m...... the map is 6.5 in.
Sum means u add the two numbers together
38 + 27 = 65
I believe you would have to multiply both 25 and 20 and what ever number you get dived by 100 if the numbers to high multiply aging or subtract the number (if it's wrong I'm really not good at my math I'm sorry)
